Related to Every normal operator on a separable Hilbert space has a square root that commutes with it

Does it exist an automorphism $f$ in a separable $\mathbb C$ Hilbert space, such that $f$ has no square root?

If so, a concrete example would be useful.

  • $\begingroup$ Certainly yes for a real Hilbert space, for example let $H=\Bbb R$ and define $T:H\to H$ by $Tx=-x$. $\endgroup$ – David C. Ullrich Nov 8 at 20:49
  • $\begingroup$ I should have mentioned over $\mathbb C$. I edited the question. Thanks David. $\endgroup$ – user384617 Nov 8 at 21:12
  • $\begingroup$ The shift operators on $l^2$ do not have square roots: math.stackexchange.com/questions/485259/… The proofs there should work for $l^2$ equally well $\endgroup$ – daw Nov 8 at 21:33
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    $\begingroup$ Thanks but these are not automorphisms $\endgroup$ – user384617 Nov 8 at 22:20
  • $\begingroup$ The shift operator on $\ell_2(\Bbb Z)$ is an automorphism. Of course it has a square root, and in fact it's normal... $\endgroup$ – David C. Ullrich Nov 9 at 1:55

Yes, such operators exist. This was proven by Halmos, Lumer, and Schäffer, Proc. AMS, 4, 1 (1953), 142-149.

Concretely, given a domain $D\subset\mathbb C$ define $$ D^{1/2}=\{\lambda\in\mathbb C:\ \lambda^2\in D\}. $$ They proved that the multiplication operator $M_z\in B(L^2(D))$ given by $(M_zf)(z)=zf(z)$ has a square root if and only if $D^{1/2}$ is disconnected. This can be seen to be equivalent to $D$ surrounding (but not containing, obviously) the origin.

So, if for instance you take any disk that does not contain the origin, say $D=\{\lambda:\ |\lambda-2|<1\}$, then $M_z\in B(L^2(D))$ is invertible and has no square root.


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